7.5 Computational experience
7.5.1 Instances
Our data comprises 9 instances that capture different aspects of real-world networks. In particular, these instances vary in size, type, number and diameter of the pipes that can be installed. Moreover, some special requirements to be discussed below are sometimes present.
The main characteristics of the instances are reported in Table 7.1.
For each instance, Table 7.1 reports the name and the numbers of junctions (including the reservoirs), reservoirs, pipes and diameter choices. Moreover, the column labeled “duplicates” indicates the number of pipes whose diameter is fixed but which can possibly be duplicated by installing a new pipe (whose diameter must be determined) in parallel. Finally, the last column indicates which currency is used to express the unit cost of the pipes, namely, US Dollar ($), Italian Lira () and Euro (e).
Instances shamir, hanoi, blacksburg and New York are taken from the literature, while the others are real-world instances of Italian water networks4.
For the instances from the literature, the only one that requires some preprocessing of the data in order to fit into our definitions is New York which will be separately discussed below. However, the data for instance blacksburg available from [116] was incomplete, and the final version of the instance that we used and make available is certainly (slightly) different from the original one.
3The percentage deviation of algorithm A with respect to algorithm B is computed as 100 × (value[A] −
value[B])/value[B].
7.5. COMPUTATIONAL EXPERIENCE 97
Table 7.1: Water Networks.
number of . . . unit
name junctions reservoirs pipes duplicates diameters cost
shamir 7 1 8 – 14 hanoi 32 1 34 – 6 blacksburg 31 1 35 – 11 New York 20 1 21 21 12 foss poly 0 37 1 58 – 7 foss iron 37 1 58 – 13 e foss poly 1 37 1 58 – 22 e pescara 71 3 99 – 13 e modena 272 4 317 – 13 e
For the real-world instances, the three instances “foss X” refer to a single neighborhood of Bologna, called Fossolo. Instance foss poly 0 consists of the original data provided to us and the pipe material for that instance is polyethylene. Instance foss iron is for the same network, but with almost twice as many choices of pipe diameters and with the material being cast iron. For instance foss poly 1 the material for the pipes is again polyethylene but there are more choices than foss poly 0 for the pipe diameters.
The cost data for foss poly 0 is out of date, and so the solution values cannot be di- rectly compared to those of foss poly 1 and foss iron, which, in turn, can be reasonably compared. The value of the solution reported in Section 7.5.2 for foss poly 1 is much lower than for foss iron. At first this seems surprising, but it is because polyethylene is much cheaper than cast-iron.
Finally, pescara and modena are reduced versions of the water distribution networks of two medium-size Italian cities. The pipe material is cast iron and both costs and diameters are up-to-date values in the Italian market.
The famous New York instance
The New York instance was first introduced by [114]. The problem we need to solve for this instance is quite different from the original one. Given an existing network, the objective is to “renovate” it by considering the increase of the water demand due to the population growth. The existing network is no longer adequate for the increased demand, resulting in pressure violations at several junctions. Thus, the network must be modified by duplicating some of the pipes, i.e., putting new pipes in parallel with the existing ones, at a minimum cost.
The decisions one has to take are:
1. select the pipes that need to be duplicated;
2. for each of these pipes, choose a diameter within the available diameter set.
In other words, with respect to our model, one has to add the null value to the diameter set: if such a null value is selected, it corresponds to the reverse of the decision 1 above, i.e., the pipe is not duplicated. However, such an explicit addition of the null diameter requires relevant modifications (consider the head loss equations), and an overall deterioration of our model.
Thus, we decided to handle such a case by an alternative method, along a line proposed by [114]. Note that this approach was not presented and formally stated in [114], but it can be read from the code reported in that paper. For the sake of clarity and completeness, we report it explicitly here.
The idea is to transform the problem, that includes the two decisions above, into our orig- inal problem, thus avoiding the first decision. We can easily do it introducing the equivalent pipe concept: We treat implicitly the two parallel pipes by means of a unique equivalent pipe that reproduces the same behavior at the extreme junctions of the pipe within the network. For each diameter of the duplicated pipe (including the null one) there is a discrete equivalent diameter associated with the pair existing/duplicated pipes.
We can prove the following simple result:
Theorem 7.2.. For each pipe e ∈ E the new diameters and costs are, respectively: Dnew(e, r) = Df ix(e) 4.87 1.852 + D(e, r) 4.87 1.852 1.852 4.87
Cnew(e, r) = C(e, r) ,
with r = 0, 1, . . . , re and where Df ix(e) is the diameter of the existing pipe and D(e, 0) = C(e, 0) = 0 .
Proof. Formally, for each existing pipe e ∈ E, we add two pipes e′and e′′corresponding to the duplicated and equivalent pipes, respectively. First, note that the flow through the existing and duplicated pipes must follow the same direction because they have the same start and end junctions and, consequently, the same hydraulic head which determines the flow direction. Thus, Q(e), Q(e′) and Q(e′′) agree in sign and denote the flows over the corresponding pipes. In order to impose the above described equivalence we must solve the following system of equations:
Q(e) + Q(e′) = Q(e′′)
H(i) − 10.7 · Q(e)1.852· k(e)−1.852· D(e)−4.87· len(e) − H(j) = 0 H(i) − 10.7 · Q(e′)1.852· k(e′)−1.852· D(e′)−4.87· len(e′) − H(j) = 0 H(i) − 10.7 · Q(e′′)1.852· k(e′′)−1.852· D(e′′)−4.87· len(e′′) − H(j) = 0 .
As required, these equations guarantee that, substituting the two parallel pipes with the equivalent one, we obtain the same flow and the same head loss at the start and end junctions.
The system above can be easily simplified by substituting out the flows: H(i) − H(j) 10.7 · len(e) 1 1.852 · k(e) · D(e)1.8524.87 + H(i) − H(j) 10.7 · len(e′) 1 1.852 · k(e′) · D(e′)1.8524.87 = H(i) − H(j) 10.7 · len(e′′) 1 1.852 · k(e′′) · D(e′′)1.8524.87 .
Because len(e) = len(e′) = len(e′′) and, in this instance, k(e) = k(e′) = k(e′′), it is easy to see that:
D(e′′) =D(e)1.8524.87 + D(e′) 4.87 1.852
1.852 4.87
, which proves the result.
7.5. COMPUTATIONAL EXPERIENCE 99